Limits of SequencesDate: 02/25/2001 at 02:11:06 From: N. D. Kapoor Subject: A question about limits Dear Dr. Math, We know that the limit (n -> infinity) [(1 + 1/n)^n] = e. Is the limit (n -> infinity) [(1 + 1/sqrt(n))^(1.5n)] also e? Also, what is the limit (n -> infinity) [(1 + a/n)^n], where a is not equal to 0? Please help. N. D. Kapoor Date: 02/26/2001 at 22:27:50 From: Doctor Fenton Subject: Re: A question about limits Dear Mr. Kapoor, Thanks for writing to Dr. Math. For questions like these, I like to use the fact that for a > 0 and b real, a^b = e^[b*ln(a)] where ln is the natural logarithm. Then: (1 + 1/sqrt(n))^(1.5n) = e^[1.5*n*ln(1 + 1/sqrt(n))] The exponent can be written as the indeterminate form: ln( 1 + 1/sqrt(n)) ------------------ 2/(3n) By L'Hospital's Rule, the limit of this expression as n->oo is the same as the limit as n->oo of: 1 1 ----------- * --------- 1 + sqrt(n) 2*sqrt(n) ------------------------- -2 ----- 3*n^2 and the quotient is of order n as n->oo, so the expression diverges to -oo, which means the original expression -> 0, not e. For the second expression, you can repeat this computation, or write: (1+a/n)^n = (1 + a/n)^[(n/a)*a] = [(1 + a/n)^(n/a)]*a = {[ 1 + 1/(n/a)]^(n/a)}^a and the term in braces {} approaches e by your earlier observation. If you have further questions, please write us again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
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